Numpy中的行交换 [英] Row exchange in Numpy

问题描述

在Python中，我可以通过多重影响来交换2个变量；它也适用于列表:

In Python I can exchange 2 variables by mean of multiple affectation; it works also with lists:

l1,l2=[1,2,3],[4,5,6]
l1,l2=l2,l1
print(l1,l2)
>>> [4, 5, 6] [1, 2, 3]

但是当我要交换2个numpy数组的行时(例如在高斯算法中)，它将失败:

But when I want to exchange 2 rows of a numpy array (for example in the Gauss algorithm), it fails:

import numpy as np
a3=np.array([[1,2,3],[4,5,6]])
print(a3)
a3[0,:],a3[1,:]=a3[1,:],a3[0,:]
print(a3)
>>> [[1 2 3]
     [4 5 6]]
    [[4 5 6]
     [4 5 6]]

我认为，出于一个奇怪的原因，这两列现在指向相同的值；但这不是事实，因为在前几行之后的a3[0,0]=5会更改a3 [0,0]，但不会更改a3 [1,0].

I thought that, for a strange reason, the two columns were now pointing to the same values; but it's not the case, since a3[0,0]=5 after the preceeding lines changes a3[0,0] but not a3[1,0].

我已经找到解决该问题的方法:例如a3[0,:],a3[1,:]=a3[1,:].copy(),a3[0,:].copy()可以工作.但是谁能解释为什么用多位行进行多种影响的交换会失败吗?我的问题与Python和Numpy的基础工作有关.

I have found how to do with this problem: for example a3[0,:],a3[1,:]=a3[1,:].copy(),a3[0,:].copy() works. But can anyone explain why exchange with multiple affectation fails with numpy rows? My questions concerns the underlying work of Python and Numpy.

推荐答案

这可以按照您希望的方式起作用:

This works the way you intend it to:

a3[[0,1]] = a3[[1,0]]

元组分配中的两个单独的分配不相互缓冲；一个接一个发生，导致覆盖您的观察

The two separate assignments in the tuple assignment are not buffered with respect to eachother; one happens after the other, leading the overwriting your observe

这篇关于Numpy中的行交换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！